Pascals Triangle

Main Question or Discussion Point

how do i find the independent of x in (6x^(2)-3/x)^6

I know that the formula nCr a^(n-r) b^r

i have got an answer of n = 6 and r = 4 i just wanna make sure that im correct and i would like someone to give me a hand~ :)

I don't think people (including myself) understand what you mean by finding "the independent of x".

HallsofIvy
Homework Helper
Do you mean possibly the "constant term" (which is "independent of x") when you expand the polynomial? If so, then yes, r= 4 is correct. However, in future, it would be good to show how you got your answer so (just in case you are wrong!) we could indicate where you went wrong.

I believe the sought expression is "coefficient". As in: Find the coefficient of x in the expansion of the expression $(6x^2-\frac{3}{x})^6$.

Obviously n = 6 here. In this case one would need to find what value of r results in $a^r b^{(n-r)}$ being a multiple of x (here $a=6x^2,b=-\frac{3}{x}$.

--Elucidus

I don't think people (including myself) understand what you mean by finding "the independent of x".
independent of x simply means that x^0 and yes I am trying to find the co-efficient there

heres are my working out.
independent of x in (6x^(2)-3/x)^6

1. (6^(6-r) x^(12-2r) ((-3^(r))/x^(r))

2. x^(12-2r)/x^(r) = x^(0)

3. x^(12-3r)=x^(0)

4. 12-3r=0

5. r = 4

independent of x simply means that x^0 and yes I am trying to find the co-efficient there

heres are my working out.
independent of x in (6x^(2)-3/x)^6

1. (6^(6-r) x^(12-2r) ((-3^(r))/x^(r))

2. x^(12-2r)/x^(r) = x^(0)

3. x^(12-3r)=x^(0)

4. 12-3r=0

5. r = 4
Ah, I see. This is a term I was unfamiliar with, but I understand now. Your work seems correct so far. If r = 4, what would the independent coefficient (constant term) be?

--Elucidus

thx for the help :)

exams finally over